Expected Fisher Information Matrix for Gamma Distribution using canonical link

How to find the fisher information matrix for a random variable $$Y \sim$$ Gamma$$(\nu,\alpha)$$?

$$0 < \nu, \alpha, y < \infty$$

I have written:

$$f_Y(y; \nu, \alpha) = \frac{y^{\nu-1}{\alpha}^{\nu}e^{-y\alpha}}{\Gamma (\nu)} \mathbb{1}_{Y \in (0, \infty)}$$

$$= \exp \{ -y\alpha + \nu \log \alpha + (\nu-1) \log y - \log \Gamma (\nu) \} \mathbb{1}_{Y \in (0, \infty)}$$

$$= \exp \{ \frac{y (- \alpha / \nu ) - [- \log \alpha ]}{1/\nu} + (\nu-1) \log y - \log \Gamma (\nu) \} \mathbb{1}_{Y \in (0, \infty)}$$

such as

$$f_Y(y; \theta, \phi) = \exp \{ \frac{y \theta - [- \log (- \theta) ]}{\phi} - \frac{\log(\phi)}{\phi} + (1/\phi -1) \log y - \log \Gamma (1/\phi) \} \mathbb{1}_{Y \in (0, \infty)}$$

where,

$$b(\theta) = - \log (- \theta)$$,

$$c(\theta) = - \frac{\log(\phi)}{\phi} + (1/\phi -1) \log y - \log \Gamma (1/\phi)$$,

$$\theta \equiv \frac{-\alpha}{\nu}$$, e

$$\phi \equiv 1/\nu$$

$$\mu \equiv E[Y] = B'(\theta) = \frac{d B(\theta)}{d \theta} = - \frac{1}{\theta} = -\theta^{-1} = \frac{\nu}{\alpha}$$

$$\text{var} [Y] = B''(\theta)\phi = \frac{d B'\theta)}{d \theta} \phi = \frac{1}{\theta^2} \phi = \phi\mu^2 = \frac{\nu}{\alpha^2}$$

And then, replacing $$\mu$$ in $$f_Y(y)$$:

$$f_Y(y) = \exp \{ \frac{y (-\frac{1}{\mu}) - \log (\mu) }{\phi} - \frac{\log(\phi)}{\phi} + (1/\phi -1) \log y - \log \Gamma (1/\phi) \} \mathbb{1}_{Y \in (0, \infty)}$$

$$g(\mu) = -\frac{1}{\mu}$$